問題描述：
輸入：按x座標升序排列的n（n>=2）個點的集合，S={(x1,y1),(x2,y2)….(xn,yn)}
輸出：最近點對的距離。
分析：
用m=（low+high）/2,講點劃分為兩部分，d1=左半部分點的最近距離，d2=右半部分點的最近對距離，d=min(d1,d2)。還需考慮在中線兩邊的點的最近距離，且只用考慮與中線距離

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

struct point
{
    int x,y;
};

int
 
 cmp(point a ,point b)
{
    return a.y<b.y;
}

double Disttance(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double Closest(point s[],int low,int high);

int n;
int main()
{
    double MinD;
    point p[10010];
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin
 
>>p[i].x>>p[i].y;
    }
    MinD = Closest(p,0,n-1);
    cout<<MinD<<endl;

    return 0;
}

double Closest(point s[],int low,int high)
{
    double d1,d2,d3,d;
    int mid,i,j,index;
    point p[n];
    if(high-low == 1)
        return Disttance(s[low],s[high]);
    if(high-low == 2
 
)
    {
        d1 = Disttance(s[low],s[low+1]);
        d2 = Disttance(s[low+1],s[high]);
        d3 = Disttance(s[low],s[high]);
        if(d1<d2 && d1<d3)
            return d1;
        else
        {
            if(d2<d3)
                return d2;
            else
                return d3;
        }
    }
    else
    {
        mid = (low+high)/2;
        d1 = Closest(s,low,mid);
        d2 = Closest(s,mid+1,high);
        if(d1<d2)
            d = d1;
        else
            d = d2;
        index = 0;
        for(i=mid;i>=low && s[i].x-s[mid].x<d;i--)
            p[index++] = s[i];
        for(j=mid+1;j<=high && s[j].x-s[mid].x<d;j++)
            p[index++] = s[j];
        sort(p,p+index,cmp);
        for(i=0;i<index;i++)
        {
            for(j=i+1;j<index;j++)
            {
                if(p[j].y-p[i].y>=d)
                    break;
                else
                {
                    d3 = Disttance(p[j],p[i]);
                    if(d3<d)
                        d = d3;
                }
            }
        }
    }
    return d;
}